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3 years ago

2 ÷3 × 6 whole number 3 on 4, into bracket 2 whole number 4 on 15 - 1 whole number 2 on 3

Mathematics

1 answer:

Aleonysh [2.5K]3 years ago

4 0

Answer:

the question is not clear

Step-by-step explanation:

wdym by this 2 ÷3 × 6 whole number 3 on 4, into bracket 2 whole number 4 on 15 - 1 whole number 2 on 3

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Mis Meyer's paid $3600 all together for the equipment, furniture and decorations for her restaurant. The equipment cost $500 mor

Tju [1.3M]

Answer:

The equipment will cost $1740.

Step-by-step explanation:

Let the Cost of equipment, furniture and decoration be x, y and z.

Now, According to question,

x + y + z = 3600 ...... (1) (cost of all items)

x = 500 + y (∵ equipment cost 500 more than furniture)

and y = 2z ( ∵ furniture twice as much as decoration)

so, z = y/2

Now substituting the value of x and z in eq (1)

$x + y + z = 3600$

$500 + y + y + \frac{y}{2} = 3600$

$2y + \frac{y}{2} = 3600 - 500$

$\frac{5y}{2} = 3100$

$y = \frac{3100\times 2}{5} = 1240$

So, the cost of furniture (y) = 1240

∴ Cost of equipment = y + 500 = 1240 + 500 = 1740

Therefore the cost of equipment was $1740.

3 0

3 years ago

Do you know what 2/9 x 3 is?

ohaa [14]

Answer:

$answer = \frac{2}{3}$

$\frac{2}{9} \times \frac{3}{1} \\ = \frac{6}{9} = \frac{2}{3}$

7 0

3 years ago

Read 2 more answers

Is 3y^2-2y already simplified

geniusboy [140]

Yes it is already simplified because there is nothing else that you are able to do to it.

4 0

3 years ago

Read 2 more answers

Help me with differentation and integration please!!

Marina86 [1]

Answer:

See below

Step-by-step explanation:

$\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x$

Recall

$\dfrac{d}{dx}\tan x=\sec^2$

Using the chain rule

$\dfrac{dy}{dx}= \dfrac{dy}{du} \dfrac{du}{dx}$

such that $u = \tan x$

we can get a general formulation for

$y = \tan^n x$

Considering the power rule

$\boxed{\dfrac{d}{dx} x^n = nx^{n-1}}$

we have

$\dfrac{dy}{dx} =n u^{n-1} \sec^2 x \implies \dfrac{dy}{dx} =n \tan^{n-1} \sec^2 x$

therefore,

$\dfrac{d}{dx}\tan^3 x=3\tan^2x \sec^2x$

Now, once

$\sec^2 x - 1= \tan^2x$

we have

$3\tan^2x \sec^2x = 3(\sec^2 x - 1) \sec^2x = 3\sec^4x-3\sec^2x$

Hence, we showed

$\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x$

================================================

For the integration,

$\int \sec^4 x\, dx$

considering the previous part, we will use the identity

$\boxed{\sec^2 x - 1= \tan^2x}$

thus

$\int\sec^4x\,dx=\int \sec^2 x(\tan^2x+1)\,dx = \int \sec^2 x \tan^2x+\sec^2 x\,dx$

and

$\int \sec^2 x \tan^2x+\sec^2 x\,dx = \int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx$

Considering $u = \tan x$

and then $du=\sec^2x\ dx$

we have

$\int u^2 \, du = \dfrac{u^3}{3}+C$

Therefore,

$\int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx = \dfrac{\tan^3 x}{3}+\tan x + C$

$\boxed{\int \sec^4 x\, dx = \dfrac{\tan^3 x}{3}+\tan x + C }$

6 0

2 years ago

HELP ME PLZ SURFACE AREA OF A RAMP

olganol [36]

Answer:

i think it 37 square feet

6 0

3 years ago